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Consider the following C program:

#include <stdio.h>

void swap(int []); 

int main(void) {
     int a[3] = {1, 2, 3};
     swap(a);
     printf("%d %d %d\n", a[0], a[1], a[2]);
     swap(&a[1]);
     printf("%d %d %d\n", a[0], a[1], a[2]);
     return 0; 
}

void swap(int x[]) {
     int temp = x[0];
     x[0] = x[1];
     x[1] = temp; 
}

Can anyone explain why the following code gives an output
2 1 3
2 3 1

rather than
2 1 3
1 2 3

I understand that swap(a) swaps a[0] and a[1]. But I'm not so sure how swap(&a[1]) works. In both cases we are passing the array a into the function swap, are we not? So my hypothesis was that swap(&a[1]) should again swap a[0] and a[1], giving us back the original order 2 3 1.

EDIT: This code was written as intended. I just wanted to see what happens if we pass an address of an element other than the first element into the function. Apparently if I pass &a[n] into the function, it disregards all elements before a[n] and treats a[n] as the first element?

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  • Yes, I understand that. But how exactly does swap(&a[1]) work? In other words, what is going on when we pass into the function an address that is not the first element? Does it then treat &a[1] as the address of the first element? Commented Nov 25, 2014 at 4:47
  • change swap(&a[1]) to swap(&a[0]) Commented Nov 25, 2014 at 4:49

4 Answers 4

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1

Your swap function takes and address of an integer as input and swaps element at that address with element at next address. 

swap(a) is same as swap(&a[0])
It will swap element at index 0 with element at index 1.
Now your array becomes {2,1,3}
swap(&a[1]) swaps element at index 1 with element at index 2.
So your array now becomes {2,3,1}

Remember when you pass an array actually you are passing address of element at index 0 (&a[0]).

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5 Comments

So when I pass &a[1] into the function, it treats &a[1] as though it was &a[0]? Because in the function definition of swap, it clearly indicates swapping the elements at index 0 and 1.
@Vizuna: When you pass an array to a function it decays to a pointer. Your function signature is the same as saying void swap(int *x). So the function assumes it has a pointer to at least two consecutive int and swaps the values at *(x+0) and *(x+1).
Yes, when you pass &a[1] x[0] is same as a[1]. Means your x (array in swap function) starts from index 1 of a (array in your main)
@Blastfurnace I see. So if I pass the pointer into the function, it treats the element pointed to as the first element, regardless of its position in the original array?
@Vizuna: Yes, you must remember that the function doesn't actually see an array type, just a pointer. That's also why you can't determine the "size" of an array parameter, it's already decayed to a pointer.
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Because when you swap first time you sent an array starting with 0 index then it exchanges 0th and 1st index values means at first swap, array becomes {2,1,3} then again you sent this array starting with 1 index means you sent only two values {1,3}, then that values gets exchanges. In this case x[0] = 1 & x[1] = 3, that is why the output came as {2,3,1} here 2 remains as it is.

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In the first call swap(a); you are passing the entire array (actually a pointer to the first element.) So you are swapping elements 0 and 1, which are 1 and 2, respectively, therefore your first output is 2 1 3.

In the second call swap(&a[1]); you are passing the address of the second element in the array, and in your swap function you are therefore swapping the second and third elements, at indices 1 and 2, which are 1 and 3, respectively. That's why your answer after the second iteration is 2 3 1.

Possibly what you meant was

swap(&a[0]);

This will swap the values at indices 0 and 1, which will give you the desired output of 1 2 3.

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In your code before the first println statement you are passing a complete array to the parameter of the fuction but in the next you are just passing a single value of location of a[1] which make bit confusing for the function because it has only a single value to swap or it might be as first you passed a zero(0) index or the array to swap with the next time you pass first 1 index of array to swap with which means [1,3 to swap] so pass complete in next statment as this 

 swap(a);
 printf("%d %d %d\n", a[0], a[1], a[2]);
 swap(a);
 printf("%d %d %d\n", a[0], a[1], a[2]);
 return 0;

this will generate your desire answer

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